Why has the Si diode been open circuited?
Let's consider if you only had the germanium diode in series with the 2.2k Ohm resistor for now. One way to solve for the voltage is to do it graphically. This is done by plotting the load line of the resistor. This load line is ploted by shorting the diode to and finding the current through the resistor is 5.5mA. To find another point on the graph the diode is open circuited and the voltage is found to be 12V. The line through (0V,5.5ma) and (12V,0mA) can be plotted plotted on this graph. The intersection of this line and the diode curve is this diodes opperating point.
Now you should see the Ge diode limits the voltage to a little more than 0.3V. So now if you put the Si diode back in the circuit, it can't have more than this 0.3V across it. The Si diode at 0.3V has zero current through it.
The answer to you question is that the Si diode has been open circuited because the Ge diode limits the voltage across it to 0.3 where it has zero current.
In finding the answer for your question look at the properties of the diodes. Maybe you know the difference between a germanium and silicium diode.
If not then try to find information concerning the voltage- current caracteristic of both diodes.
You can find this information in Wikipedia or the datasheets of germanium and silicium diodes.
Once you have this you should be able to answer the question yourself.
Note resulting from the comments:
The voltagedrop across a diode is also the threshold for conducting or not.
In other words it is not only a voltagedrop as if you are looking at two resistors in parallel. If you look again at the datasheets and you compare the voltages during the conducting state then you should be able to determine why the silicon diode remains open.